\documentclass{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

\usepackage{cancel}
\usepackage{graphicx}
\usepackage{hyperref}

\title{Problem Solution}
\author{Hu Jingcheng 3180105323@zju.edu.cn}

\begin{document}
\maketitle
\section{Problem 1.1}
First we draw the simplified diagram for the circuits:
\begin{figure}[h]
    \centering
    \includegraphics[scale=0.3]{pics/circuits.jpg}
    \caption{Simplified Diagram for Circuits}
\end{figure}

Then we can write down Lagrangian directly:
\begin{align}
    L&= \frac{C\dot\Phi_1^2}2 + \frac{C\dot\Phi_2^2}2+ \frac{C_c{(\dot\Phi_2-\dot\Phi_1)}^2}2 - \frac{{\Phi_1}^2}{L_j}-\frac{{\Phi_2}^2}{L_j}\\
     &=\frac{(C+C_c)\dot\Phi_1^2}2+\frac{(C+C_c)\dot\Phi_2^2}2 - C_c\dot\Phi_1\dot\Phi_2- \frac{{\Phi_1}^2}{L_j}-\frac{{\Phi_2}^2}{L_j}
\end{align}
Then write down the canonical momentum:
\begin{align}
    \frac{\partial L}{\partial \dot\Phi_1} &= (C+C_c)\dot\Phi_1 - C_c \dot\Phi_2 = q_1\\
    \frac{\partial L}{\partial \dot\Phi_2} &= - C_c \dot\Phi_1+(C+C_c)\dot\Phi_2  = q_2
\end{align}
Then use $q_1,q_2$ to solve $\dot\Phi_1,\dot\Phi_2$:
\begin{align}
   \begin{bmatrix} \dot\Phi_1 \\ \dot\Phi_2 \end{bmatrix} 
   = 
   \frac1{C(C+2C_c)}
   \begin{bmatrix}
       (C+2C_c)q_1 + C_c q_2\\
       C_c q_1+ (C+2C_c)q_2 
   \end{bmatrix}
   \implies 
   \dot\Phi_2-\dot\Phi_1 = \frac1{C+2C_c}(q_1-q_2)
\end{align}

Then we use the relation $H = p_i q_i - L$, and eliminate $\dot\Phi_i$, then we get the Hamiltonian:
\begin{align}
H = \frac{{\Phi_1}^2}{L_j}+\frac{{\Phi_2}^2}{L_j} + \frac{C+C_c}{C+2C_c}\frac{q_1^2}{2C}+\frac{C+C_c}{C+2C_c}\frac{q_2^2}{2C} + \frac{C_c}{{(C+2C_c)}^2}q_1q_2
\end{align}
In the limit that $C_c \ll C$, we have:
\begin{align}
    H \approx \frac{{\Phi_1}^2}{L_j}+\frac{{\Phi_2}^2}{L_j} + \frac{q_1^2}{2C}+\frac{q_2^2}{2C} + \frac{C_c}{C^2}q_1q_2
\end{align}
where we can also see that this is in the weak coupling limit.
Then we can reexpress the coupling Hamiltonian in each raising and lowering operators:
\begin{align}
    H_i &= -\frac{C_c}{C}\omega_0 (c_1 - c_1^\dagger)(c_2 - c_2^\dagger)\\
        &= \frac{C_c}{C}\omega_0(c_1c_2^\dagger + c_1^\dagger c_2)
\end{align}

\end{document}